The lost property office at a mainline station records items left on trains handed in in two categories: Category A is clothing and Category B is paperwork.
Over a period of time the mean number of items in category A handed in per day is $2$ and the mean number in category B per day is $3$.
These distributions may be assumed to be independent of each other and each may be modeled by the Poisson distribution.
Find the probability that
- in one day there are no items handed in;
- in one day there is exactly one item handed in;
- in one day there are at most 2 items handed in.
Mean number of items in category A handed in per day $= 2$
Mean number of items in category B handed in per day $= 3$
The number of items handed in per day, in category A and category B, are both modeled by the Poisson distribution.
$\therefore \;$ Parameter of Poisson distribution for both categories together $= \lambda = 2 + 3 = 5$
- $P\left(\text{No items are handed in}\right)= P \left(X = 0\right)$
$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 5} = 0.0067 \end{aligned}$
- $P \left(\text{exactly 1 item handed in}\right) = P \left(X = 1\right)$
$\begin{aligned} P \left(X = 1\right) & = \dfrac{e^{- \lambda} \times \lambda^1}{1!} \\\\ & = e^{- \lambda} \times \lambda \\\\ & = e^{- 5} \times 5 \\\\ & = 0.0067 \times 5 = 0.0335 \end{aligned}$
- $P \left(\text{at most 2 items handed in}\right) = P \left(X \leq 2\right)$
$\begin{aligned} P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 5} \times \left[1 + \dfrac{5}{1} + \dfrac{5^2}{2} \right] \\\\ & = 0.0067 \times 18.5 = 0.1240 \end{aligned}$